In the figure, EFGH is a quadrilateral where ∠EFG = 126° and ∠FGH = 114°. ∠GHJ = 72° and FG = GH. The point J on EH is such that FJ is parallel to GH. Calculate
- ∠FHJ
- ∠FEJ
(a)
∠GHF
= (180° - 114°) ÷ 2
= 33° (Isosceles triangle)
∠FHJ
= 72° - 33°
= 39°
(b)
∠FEJ
= 360° - 114° - 72° - 126°
= 48° (Sum of angles in a quadrilateral)
Answer(s): (a) 39°; (b) 48°